Question

One object is traveling northeast with a momentum of 12 kg-m/s, while a second object is traveling southwest on the same path as the first object with a momentum of 4 kg-m/s. When the two objects collide, they stick together. What is the momentum after their collision?(1 point)
Responses

8 kg-m/s northeast
8 kg-m/s northeast

8 kg-m/s southwest
8 kg-m/s southwest

16 kg-m/s northeast
16 kg-m/s northeast

16 kg-m/s southwest
16 kg-m/s southwest

Answer 1

To determine the momentum after the collision, we need to consider the momentum vectors of both objects.

1. The first object is traveling northeast with a momentum of \(12 \, \text{kg-m/s}\).
2. The second object is traveling southwest with a momentum of \(4 \, \text{kg-m/s}\).

To find the total momentum after the collision, we can break this down into components:

- **Northeast momentum (first object):**
- This can be split into x and y components (assuming northeast corresponds to a 45-degree angle from both axes):
- \( p_{1x} = 12 \cos(45^\circ) = 12 \cdot \frac{1}{\sqrt{2}} = 6\sqrt{2} \, \text{kg-m/s} \)
- \( p_{1y} = 12 \sin(45^\circ) = 12 \cdot \frac{1}{\sqrt{2}} = 6\sqrt{2} \, \text{kg-m/s} \)

- **Southwest momentum (second object):**
- This also corresponds to a 45-degree angle but in the opposite direction:
- \( p_{2x} = -4 \cos(45^\circ) = -4 \cdot \frac{1}{\sqrt{2}} = -2\sqrt{2} \, \text{kg-m/s} \) (negative because it's moving in the negative x direction)
- \( p_{2y} = -4 \sin(45^\circ) = -4 \cdot \frac{1}{\sqrt{2}} = -2\sqrt{2} \, \text{kg-m/s} \) (negative because it's moving in the negative y direction)

Now, we can find the total momentum components in the x and y directions:

1. **Total x-component of momentum:**
\[
p_{x, \text{total}} = 6\sqrt{2} + (-2\sqrt{2}) = (6 - 2)\sqrt{2} = 4\sqrt{2} \, \text{kg-m/s}
\]

2. **Total y-component of momentum:**
\[
p_{y, \text{total}} = 6\sqrt{2} + (-2\sqrt{2}) = (6 - 2)\sqrt{2} = 4\sqrt{2} \, \text{kg-m/s}
\]

Now we can find the magnitude of the total momentum vector:
\[
p_{\text{total}} = \sqrt{(p_{x, \text{total}})^2 + (p_{y, \text{total}})^2} = \sqrt{(4\sqrt{2})^2 + (4\sqrt{2})^2} = \sqrt{32 + 32} = \sqrt{64} = 8 \, \text{kg-m/s}
\]

To find the direction, since both components are positive, the direction of the total momentum is northeast.

Thus, the momentum after their collision is:

**8 kg-m/s northeast**.